CALCULUS 2 LECTURE VECTORS Overview In this lesson, we will determine that a vector is a quantity that has both a magnitude (or size) and a direction. Both of these properties must be given in order to specify a vector completely. Thus, we can describe how to write down vectors, how to add and subtract them, and how to use them in geometry and in calculus as well. Learning objectives At the end of the lesson, the student shall be able to: 1. Define vectors and scalars 2. Perform algebraic operations involving vectors. 3. Graphically represent vectors in two or three dimensions Discussion Scalars and Vectors Quantities of time, mass, temperature and speed are scalar quantities. These quantities has magnitude only and denoted by numbers. Vector quantities are quantities measured with magnitude and direction. In Physics examples of these quantities are force, velocity and acceleration. Vectors are represented by an arrow or a directed line segment. The direction of the vector is measured by the angle which it makes from a fixed line while the length of the arrow represents the magnitude of the vector. A vector can be indicated by printing a letter in boldface ( v ) or by putting an arrow above it. (π£β). The magnitude of a vector π ππ πΆπ· is denoted by |π| ππ |πΆπ·|. π π π −π π π πΆ Two vectors a and b are called equal π = π, if they have the same magnitude and the same direction. A vector whose magnitude is the same as that of a but in opposite direction of that of a is said to be negative of a or −π . Components of a Vector/ Unit Vector ο· All vectors can be broken down into their component vectors. ο· All component vectors can be represented by unit vectors. Two Dimensional Vector Considering a vector v (ππ , ππ ) (π₯2 , π¦2 ) y |π| π½ π¦2 − π¦1 ππ π£π₯ (π₯1 , π¦1 ) π₯2 − π₯1 Let π = β¨π―π± , π―π² β©. π―π± − πππ ππππππππππ πππππππππ ππ π π―π² − πππ ππππππππ πππππππππ ππ π |π| − ππππππππ π ππ ππππππ ππ π ⟨ ⟩ − πππππ ππππππππ π½ − πππ π ππππππππ πππππ πππ π 2 If there are ππ and ππ , using Pythagorean Theorem |π―|=√ππ π + ππ π length of two dimensional vector (ππ , ππ ) |π| ππ ππ π»πππ½ = π π ππ π½ = πππ−π (π ) π½ π ππ ππ = π πππ π½ ππ = π πππ π½ π = ππ π + ππ π ( πΆπππ‘ππ πππ ππππ‘ππ πππ‘ππ‘πππ) Illustration 1: Find the vector components and unit vector of A at 60 degree angle from positive x – axis. π΄ = 6 , π = 60° π΄π₯ = 6 cos 60° π΄ j π΄π¦ 1 √3 = 6 (2) = 6( 2 ) π΄π₯ = 3 60° i π΄π¦ = 6 sin 60° π¨π π΄π¦ = 3√3 π¨ = ππ + π√ππ Cartesian vector notation π¨ = ⟨π, π√π⟩ Component form πΌπππ ππππππ , ππ¨ (Always = 1) π π ββ π¨ π π ππ+π√ππ ππ¨ = √π π Checking: ππ¨ = √(π) + ( π ) ππ¨ = |π¨| π π ππ¨ = π π + √π π √π π Checking: ππ¨ = √(π) + ( π ) π π π π = √π + π = √π π =π Illustration 2: If π― = β¨−3, 5β©, find βπ―β and π, where 0 ≤ π < 360°. Solution : π― = β¨−3, 5β© 3 βπ―β = √(−3)2 + (5)2 = √9 + 25 = √34 π πππ π½ = π π πππ π½ = −π π½ = π¨πππππ π −π π½ = −ππ° πΈ π°π½ Three Dimensional Vector If π΅ = ⟨3, 4 , −2⟩ component form π΅ = 3π + 4π − 2π§ Cartesian vector notation |π΅| = √π΅π₯ 2 + π΅π¦ 2 + π΅π§ 2 length of the three dimensional vector = √(3)2 + (4)2 + (−2)2 = √9 + 16 + 4 |π΅| = √29 Unit Vector βββ π© ππ¨ = |π©| = ππ¨ = π √ππ 3π+4π−2π§ √ππ π+ π √ππ π− π √ππ π unit vector Coordinate Direction Angles z πΌ = πππ −1 ( 3 π½ = πππ −1 ( 4 πΎ = πππ −1 ( −2 ) = 56. .15° √29 ) = 42.03° √29 ) = 111.80° √29 B πΎ πΌ π½ y B’ x 4 Checking ; πππ 2 πΌ + πππ 2 π½ + πππ 2 πΎ = 1 πππ 2 56.14° + πππ 2 42.03° + πππ 2 111.80° = 1 0.31 + 0.55 + 0.14 = 1 1=1 Vector Addition If there are vectors a and b, their sum or resultant can be determined through the following methods: 1. Graphical Method a. Triangle Method – Connect the head of the first to the tail of the second. Then, connect the tail of the first to the head of the last. Illustration: a b b a π+π b. Parallelogram Method The vectors are drawn from the same initial point. Then, vectors parallel to these vectors will be drawn. Connecting the initial point to the intersection of the vectors parallel to the original will give the sum or resultant vector. Illustration: a b a a+b b 2. Theoretical Method a. The use of Cosine Law and Sine Law Illustration: Find the sum ( resultant) of the given vectors and the angle which it makes with the horizontal axis. 5 A = 10 units B = 7 units 140° 30° Solution: Form the triangle, R 7u πΌ 140° π 30° π½ 10 u 30° πΌ = 180° − 140° = 40° π = 30° + 40° = 70° Use Cosine Law ππ = ππ + ππ − πππ πππ πͺ πΉπ = π¨π + π©π − ππ¨π© ππππ πΉ = √π¨π + π©π − ππ¨π© ππππ πΉ = √πππ + ππ − π(ππ)(π) πππ70° πΉ = ππ. ππ π Using Sine Law πΊππ π¨ πΊππ π© πΊππ πͺ = = π π π π πππ sin 70° = 7 10.06 7 sin 70° π πππ = 10.06 π πππ = 0.65386166 π½ = πππ−π (0.65386166 ) π½ = ππ. ππ° π· = ππ° + ππ. ππ° = ππ. ππ° b. Vector Addition (Cartesian Vector notation/ Component Form) To add and subtract vectors, add or subtract (in order) the components of the vectors. 6 7
